Problem: As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him. 

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Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.

A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:

  * Empty string is a correct bracket sequence. 
  * if s is a correct bracket sequence, then (s) is also a correct bracket sequence. 
  * if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence. 



A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.

Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.

Joyce doesn't know anything about bracket sequences, so she asked for your help.

Input

The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).

Output

Print the answer to Will's puzzle in the first and only line of output.

Examples

Input

((?))


Output

4


Input

??()??


Output

7

Note

For the first sample testcase, the pretty substrings of s are:

  1. "(?" which can be transformed to "()". 
  2. "?)" which can be transformed to "()". 
  3. "((?)" which can be transformed to "(())". 
  4. "(?))" which can be transformed to "(())". 



For the second sample testcase, the pretty substrings of s are:

  1. "??" which can be transformed to "()". 
  2. "()". 
  3. "??()" which can be transformed to "()()". 
  4. "?()?" which can be transformed to "(())". 
  5. "??" which can be transformed to "()". 
  6. "()??" which can be transformed to "()()". 
  7. "??()??" which can be transformed to "()()()". 

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.util.stream.Stream;
import java.util.Map.Entry;

public class Main implements Runnable {
    static final int MOD = (int) 1e9 + 7;
    static final int MI = (int) 1e9;
    static final long ML = (long) 1e18;
    static final Reader in = new Reader();
    static final PrintWriter out = new PrintWriter(System.out);
    StringBuilder answer = new StringBuilder();

    public static void main(String[] args) {
        new Thread(null, new Main(), "persefone", 1 << 32).start();
    }

    @Override
    public void run() {
        solve();
        printf();
        flush();
    }

    void solve() {
        char[] s = in.next().toCharArray();
        int n = s.length;
        int ans = 0;
        for (int i = 0; i < n; i++) {
            int balance = 0, questions = 0;
            for (int j = i; j < n; j++) {
                if (s[j] == '?') {
                    questions++;
                } else {
                    balance += s[j] == '(' ? 1 : -1;
                }
                if (questions > balance) {
                    questions--;
                    balance++;
                }
                if (questions + balance < 0) break;
                if ((j - i) % 2 == 1 && questions == balance) ans++;
            }
        }
        printf(ans);
    }

    void printf() {
        out.print(answer);
    }

    void close() {
        out.close();
    }

    void flush() {
        out.flush();
    }

    void printf(Stream<?> str) {
        str.forEach(o -> add(o, " "));
        add("\n");
    }


    void printf(Object... obj) {
        printf(false, obj);
    }

    void printfWithDescription(Object... obj) {
        printf(true, obj);
    }


    private void printf(boolean b, Object... obj) {

        if (obj.length > 1) {
            for (int i = 0; i < obj.length; i++) {
                if (b) add(obj[i].getClass().getSimpleName(), " - ");
                if (obj[i] instanceof Collection<?>) {
                    printf((Collection<?>) obj[i]);
                } else if (obj[i] instanceof int[][]) {
                    printf((int[][]) obj[i]);
                } else if (obj[i] instanceof long[][]) {
                    printf((long[][]) obj[i]);
                } else if (obj[i] instanceof double[][]) {
                    printf((double[][]) obj[i]);
                } else printf(obj[i]);
            }
            return;
        }

        if (b) add(obj[0].getClass().getSimpleName(), " - ");
        printf(obj[0]);
    }

    void printf(Object o) {
        if (o instanceof int[])
            printf(Arrays.stream((int[]) o).boxed());
        else if (o instanceof char[])
            printf(new String((char[]) o));
        else if (o instanceof long[])
            printf(Arrays.stream((long[]) o).boxed());
        else if (o instanceof double[])
            printf(Arrays.stream((double[]) o).boxed());
        else if (o instanceof boolean[]) {
            for (boolean b : (boolean[]) o) add(b, " ");
            add("\n");
        } else
            add(o, "\n");
    }

    void printf(int[]... obj) {
        for (int i = 0; i < obj.length; i++) printf(obj[i]);
    }

    void printf(long[]... obj) {
        for (int i = 0; i < obj.length; i++) printf(obj[i]);
    }

    void printf(double[]... obj) {
        for (int i = 0; i < obj.length; i++) printf(obj[i]);
    }

    void printf(boolean[]... obj) {
        for (int i = 0; i < obj.length; i++) printf(obj[i]);
    }

    void printf(Collection<?> col) {
        printf(col.stream());
    }

    <T, K> void add(T t, K k) {
        if (t instanceof Collection<?>) {
            ((Collection<?>) t).forEach(i -> add(i, " "));
        } else if (t instanceof Object[]) {
            Arrays.stream((Object[]) t).forEach(i -> add(i, " "));
        } else
            add(t);
        add(k);
    }


    <T> void add(T t) {
        answer.append(t);
    }

    static class Reader {
        private BufferedReader br;
        private StringTokenizer st;

        Reader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        Reader(String fileName) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(fileName));
        }

        boolean isReady() throws IOException {
            return br.ready();
        }

        String next() {
            try {
                while (st == null || !st.hasMoreTokens()) {
                    st = new StringTokenizer(br.readLine());
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] nextIntArray(int n) {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++)
                arr[i] = nextInt();
            return arr;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        String nextLine() {
            String s = "";
            try {
                s = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return s;
        }
    }
}